Question: $ \int_{-1}^3 \int_{-\sqrt{y + 1}}^{\sqrt{y + 1}} dx \, dy$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_1^3 \int_{(x - 1)^2}^2 dy \, dx$ (Choice B) B $ \int_1^3 \int_{(x - 1)^2 - 1}^1 dy \, dx$ (Choice C) C $ \int_{-3}^3 \int_{x^2 - 1}^8 dy \, dx$ (Choice D) D $ \int_{-2}^2 \int_{x^2 - 1}^3 dy \, dx$
The first step whenever we want to switch bounds is to sketch the region of integration that we're given. Here, we see $-\sqrt{y + 1} < x < \sqrt{y + 1}$ and $-1 < y < 3$. Therefore: ${2}$ ${\llap{-}2}$ ${2}$ ${4}$ ${\llap{-}2}$ $y$ $x$ Because we're switching bounds to $dy \, dx$, we need to start with numeric bounds for $x$. We see that $-2 < x < 2$. Then we can define $y$ in terms of $x$. Thus, $x^2 - 1 < y < 3$. We want to pay attention especially to how this $y$ bound works at the edge of the $x$ bound. For example, at $x = 2$, the $y$ bound makes $y = 3$ as expected. In conclusion, the double integral after switching bounds is: $ \int_{-2}^2 \int_{x^2 - 1}^3 dy \, dx$